[fortran] huge() vs IBM extended double

[Alan Modra]

:ADDPATCH fortran:

huge() currently returns +Inf on targets using IBM extended long
doubles.  Not understanding fortran's "model numbers", I originally
submitted a patch to correct this by making huge() return the same value
as C's __LDBL_MAX__.  Apparently the right fix is to reduce emax by
one.  See
http://gcc.gnu.org/ml/fortran/2005-09/msg00556.html
http://gcc.gnu.org/ml/fortran/2005-09/msg00562.html
http://gcc.gnu.org/ml/fortran/2005-09/msg00564.html

Bootstrapped and regression tested powerpc64-linux, with 128-bit long
doubles enabled.

	PR fortran/24096
	* trans-types.c (gfc_init_kinds): Use one less for max_exponent
	of IBM extended double format.

Index: gcc/fortran/trans-types.c
===================================================================
--- gcc/fortran/trans-types.c	(revision 106520)
+++ gcc/fortran/trans-types.c	(working copy)
@@ -192,6 +192,15 @@
       gfc_real_kinds[r_index].digits = fmt->p;
       gfc_real_kinds[r_index].min_exponent = fmt->emin;
       gfc_real_kinds[r_index].max_exponent = fmt->emax;
+      if (fmt->pnan < fmt->p)
+	/* This is an IBM extended double format made up of two IEEE
+	   doubles.  The value of the long double is the sum of the
+	   values of the two parts.  The most significant part is
+	   required to be the value of the long double rounded to the
+	   nearest double.  If we use emax of 1024 then we can't
+	   represent huge(x) = (1 - b**(-p)) * b**(emax-1) * b, because
+	   rounding will make the most significant part overflow.  */
+	gfc_real_kinds[r_index].max_exponent = fmt->emax - 1;
       gfc_real_kinds[r_index].mode_precision = GET_MODE_PRECISION (mode);
       r_index += 1;
     }

-- 
Alan Modra
IBM OzLabs - Linux Technology Centre